我知道这个问题之前已经被问过好几次了，但是我找不到我的方法的错误。

问题:mysql_num_rows 返回错误结果时

$sql = "SELECT * FROM $topic WHERE $names LIKE '%$q%'";

但是如果我用以下任何一个替换 $sql，它会返回 true。

$sql = "SELECT * FROM $topic WHERE $names LIKE '%j%'";
$sql = "SELECT * FROM $topic WHERE $names ='Jack'";
$sql = "SELECT * FROM $topic WHERE $names = '$q' ";

var_dump 的结果

string(8) "Cust_Reg" string(5) "fName" string(2) "j " 
resource(8) of type (mysql result) 
string(46) "SELECT * FROM Cust_Reg WHERE fName LIKE '%j %'"

如果我更改 $sql = "SELECT * FROM $topic WHERE $names LIKE '%j%'";

$row['ID'] 的 var_dump 将显示

"SELECT * FROM Cust_Reg WHERE fName LIKE '%j%'" string(4) "NjA=" 
string(4) "NjE=" string(4) "NjQ=" string(4) "NjY=" string(4) "ODI="

如果您能纠正我的错误，我将不胜感激。

    $q = mysql_real_escape_string($_GET['search']);
    $q = strtolower($q);
    $topic = mysql_real_escape_string($_GET['test']);
    $names = mysql_real_escape_string($_GET['name']);

    // SELECT * from Account_Reg where Account_Name LIKE '%$q%'
    $table = "<table style='width:400px; padding:10; display:block;'><tbody>            
                <tr><td>ID</td><td>Account</td><td>First Name</td><td>Email</td></tr>"; 

        $sql = "SELECT * FROM $topic WHERE $names LIKE '%{$q}%'";
        $result = mysql_query($sql) or die (mysql_error());

        var_dump($topic);
        var_dump($names);
        var_dump($q);   
        var_dump($result);
        var_dump($sql); 

    if(is_resource($result) && mysql_num_rows($result) > 0){

        while($row = mysql_fetch_array($result)) {                              
            $table .= "<tr><td>".$row['ID']."</td>";
            $cryt = base64_encode($row['ID']);
            $row['ID'] = htmlspecialchars($cryt);
            $link = "profile.cust.update.php?id=". urlencode($row['ID']);
            $link = htmlentities($link);    

            if($names == "fName"){      
                $name = $row['fName'];

            }elseif($names == "Account_Name"){
                $name = $row['Account_Name'];
                $row['email_add'] = "";     
            }   

            $table .="<td></td><td><a href='" .$link ."'</a>"  .$name."</td><td>".$row['email_add']."</td></a> </tr>";
        }$table .="</tbody</table";
  }else{$table = "No row is selected"; }

最佳答案

尝试使用:

$sql = "SELECT * FROM $topic WHERE $names LIKE '%{$q}%'";

或

$sql = "SELECT * FROM $topic WHERE $names LIKE '%" . $q . "%'";

为了调试，在执行之前尝试输出$sql，看看变量是如何被替换的

echo $sql; die();

来自文档的@AdamWaite:

http://dev.mysql.com/doc/refman/5.0/en/pattern-matching.html

MySQL provides standard SQL pattern matching as well as a form of pattern matching based on extended regular expressions similar to those used by Unix utilities such as vi, grep, and sed.

SQL pattern matching enables you to use “_” to match any single character and “%” to match an arbitrary number of characters (including zero characters). In MySQL, SQL patterns are case-insensitive by default. Some examples are shown here. You do not use = or <> when you use SQL patterns; use the LIKE or NOT LIKE comparison operators instead.

关于php - 带有百分号的 MySQL/PHP LIKE 不起作用，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7242500/